GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 24 Sep 2018, 20:23

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

In the diagram above, angle C = 90º and AC = BC. Point M is the midpo

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49431
In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

Show Tags

New post 11 Mar 2015, 04:56
3
3
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

70% (03:27) correct 30% (03:31) wrong based on 156 sessions

HideShow timer Statistics

Attachment:
cpotg_img10.png
cpotg_img10.png [ 12.13 KiB | Viewed 3002 times ]
In the diagram above, angle C = 90º and AC = BC. Point M is the midpoint of AB. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The shaded region between the two arcs is called a “lune.” What is the ratio of the area of the lune to the area of triangle ABC?

A. 1
B. \(\frac{\pi}{2}\)
C. \(\frac{\pi}{3}\)
D. \(\frac{3\pi}{8}\)
E. \(\frac{4\pi}{9}\)

Kudos for a correct solution.

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
User avatar
Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 214
Location: India
MISSION : 800
WE: Design (Manufacturing)
GMAT ToolKit User Premium Member
Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

Show Tags

New post 11 Mar 2015, 05:58
2
Hi
A very time consuming
have edited the pic for all the forum members
Answer in the pic attached

cheers..!!
Attachments

lune.png
lune.png [ 28.6 KiB | Viewed 2797 times ]


_________________

Thank you

+KUDOS

> I CAN, I WILL <

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 49431
Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

Show Tags

New post 15 Mar 2015, 21:23
Bunuel wrote:
Image
In the diagram above, angle C = 90º and AC = BC. Point M is the midpoint of AB. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The shaded region between the two arcs is called a “lune.” What is the ratio of the area of the lune to the area of triangle ABC?

A. 1
B. \(\frac{\pi}{2}\)
C. \(\frac{\pi}{3}\)
D. \(\frac{3\pi}{8}\)
E. \(\frac{4\pi}{9}\)

Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This problem is based on a famous theorem of Hippocrates of Chios (c. 470 – c. 410 BCE), a predecessor of Euclid. The easiest way to see this is as follows.

Let AC = BC = 2S. This is the radius of the larger circle, which has an area of 4pi*S^2, so that the quarter circle:
Image
must have an area of pi*S^2. Hold that thought.

Now, notice that triangle ABC is a 45-45-90 triangle, so that the length of the hypotenuse must be
\(AB=2\sqrt{2}S\)
and the radius of the smaller circle is
\(AM=\sqrt{2}S\)

The area of that whole circle would be 2pi*S^2, and the area of the semicircle:
Image

would be half of that, pi*S^2. Thus, the quarter circle of the larger circle and semicircle of the smaller circle have the same area. That’s a very big idea!

Now, look at the circular segment:
Image


We don’t need to know that area: simply call it J.

If we subtract the circular segment, J, from the quarter circle of the bigger circle, we get the triangle:
pi*S^2 – J = area of triangle ABC

If we subtract the circular segment, J, from the semicircle of the smaller circle, we get the lune:
pi*S^2 – J = area of the lune

Because those two areas equal the same thing, they must equal each other.
area of triangle ABC = area of the lune

The two are equal, so their ratio is 1.

Answer = (A)
Image
Image

Attachment:
cpotg_img32.png
cpotg_img32.png [ 8.42 KiB | Viewed 2854 times ]

Attachment:
cpotg_img35.png
cpotg_img35.png [ 8.33 KiB | Viewed 2858 times ]

Attachment:
cpotg_img36.png
cpotg_img36.png [ 8.18 KiB | Viewed 2859 times ]

Attachment:
cpotg_img38.png
cpotg_img38.png [ 24.99 KiB | Viewed 2867 times ]

Attachment:
cpotg_img37.png
cpotg_img37.png [ 23.53 KiB | Viewed 2859 times ]

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior CR Moderator
User avatar
V
Status: Long way to go!
Joined: 10 Oct 2016
Posts: 1385
Location: Viet Nam
GMAT ToolKit User Premium Member
Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

Show Tags

New post 21 Nov 2016, 20:03
Attachment:
cpotg_img10.png
cpotg_img10.png [ 13.62 KiB | Viewed 1599 times ]


Let set \(AC=BC=x\)

We have \(S_{ABC}=\frac{x^2}{2}\)

\(AXBC\) is a quarter circle, so \(S_{AXBC}=\frac{\pi x^2}{4}\)

Hence \(S_{AMBX}=S_{AXBC}-S_{ABC}=\frac{\pi x^2}{4}-\frac{x^2}{2}\)

We have \(AB=x \sqrt{2}\) so \(AM=MB=\frac{x}{\sqrt{2}}\)

\(AYBM\) is a semi-circle, so \(S_{AYBM}=\frac{1}{2} \times \pi \times ( \frac{x}{\sqrt{2}} )^2=\frac{\pi x^2}{4}\)

Hence \(S_{AYBX}=S_{AYBM}-S_{AMBX}=\frac{\pi x^2}{4} - (\frac{\pi x^2}{4}-\frac{x^2}{2})=\frac{x^2}{2}\)

This means \(S_{AYBX}=S_{ABC}\).

The answer is A
_________________

Actual LSAT CR bank by Broall

How to solve quadratic equations - Factor quadratic equations
Factor table with sign: The useful tool to solve polynomial inequalities
Applying AM-GM inequality into finding extreme/absolute value

New Error Log with Timer

Senior Manager
Senior Manager
User avatar
G
Joined: 29 Dec 2017
Posts: 383
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GPA: 3.25
WE: Marketing (Telecommunications)
Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo  [#permalink]

Show Tags

New post 22 Aug 2018, 11:07
Don't mess with algebra. Assume that AC=BC=1. Calculations will be much easier and faster.
GMAT Club Bot
Re: In the diagram above, angle C = 90º and AC = BC. Point M is the midpo &nbs [#permalink] 22 Aug 2018, 11:07
Display posts from previous: Sort by

In the diagram above, angle C = 90º and AC = BC. Point M is the midpo

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.